in the given figure, ABCD is a square. if angle PQR=90 and PB=QC=DR, prove that QB=RC, PQ=QR and angle QPR=45
(a)
ABCD is a square
BC = CD
Given QC = RD
∴ BC - QC = CD - RD
⇒ BQ = CR
(b)
In ∆PBQ and ∆QCR
PB = QC [given]
BQ = CR [Proved in (a)]
∠PBQ = ∠QCR [Each is 90°]
∴ ∆PBQ and ∆QCR [SAS Congruency]
⇒ PQ = QR, ∠BPQ = ∠CQR, ∠BQP = ∠CRQ [C.P.C.T.]
(c)
BQC is a straight line
∴ ∠BQP + ∠PQR + ∠CQR = 180°
⇒∠BQP + ∠PQR + ∠BPQ = 180°
⇒(∠BQP + ∠BPQ) + ∠PQR = 180°
⇒180° - ∠PBQ + ∠PQR = 180° [Angle sum property for ∆PBQ]
⇒180° - 90° + ∠PQR = 180°
⇒∠PQR = 90°
(d)
∆PQR is an isosceles right angled triangle
∴ ∠QPR = 45°
Cheers!